Count Set Bits Brian Kernighan Algorithm in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the count variable to zero.

DSA Rust

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = [1];
    while n > 0 {
        n &= n - 1;
        count += 1;
    }
    count
}

Drag options to blanks, or click blank then click option'

An

B1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing count with 1 instead of 0.

Using n as initial count.

Starting count with a negative number.

The count variable starts at zero because we haven't counted any set bits yet.

2fill in blank

medium

Complete the code to update n by removing the rightmost set bit.

DSA Rust

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = 0;
    while n > 0 {
        n = n [1] n - 1;
        count += 1;
    }
    count
}

Drag options to blanks, or click blank then click option'

A^

B+

C|

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '&'.

Using bitwise OR or XOR which do not clear bits.

Forgetting to update n.

The '&' operator removes the rightmost set bit by doing n & (n-1).

3fill in blank

hard

Fix the error in the loop condition to correctly count set bits.

DSA Rust

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = 0;
    while n [1] 0 {
        n = n & (n - 1);
        count += 1;
    }
    count
}

Drag options to blanks, or click blank then click option'

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which stops immediately if n is not zero.

Using '<' which is incorrect for unsigned integers.

Using '!=' which also works but is less clear.

The loop continues while n is greater than zero to count all set bits.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA Rust

fn [1](mut n: u32) -> u32 {
    let mut count = 0;
    while n > 0 {
        n = n & (n - 1);
        count += 1;
    }
    [2]
}

Drag options to blanks, or click blank then click option'

Acount_set_bits

Bn

Ccount

Dreturn

Attempts:

3 left

💡 Hint

Common Mistakes

Returning n instead of count.

Using wrong function name.

Missing return statement.

The function is named count_set_bits and returns the count variable.

5fill in blank

hard

Fill all three blanks to create a test call and print the result.

DSA Rust

fn main() {
    let number = 29;
    let result = [1](number);
    println!("Number of set bits in {} is: {}", [2], [3]);
}

Drag options to blanks, or click blank then click option'

Acount_set_bits

Bnumber

Cresult

Dcount_bits

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function name in call.

Mixing variable names in print.

Forgetting to pass the number.

Call count_set_bits with number, then print number and result.

Practice

1. What does the Brian Kernighan algorithm do when counting set bits in a number?

easy

A. It repeatedly clears the least significant set bit until the number becomes zero

B. It counts the total number of bits in the number

C. It shifts the number right until it becomes zero

D. It converts the number to a string and counts '1' characters

Solution

1. Step 1: Understand the algorithm's main operation

   The Brian Kernighan algorithm works by clearing the least significant set bit in each iteration.

2. Step 2: Effect of clearing bits

   Each time the least significant set bit is cleared, the count of set bits increases by one until the number becomes zero.

3. Final Answer:

   It repeatedly clears the least significant set bit until the number becomes zero -> Option A

4. Quick Check:

   Clearing bits one by one counts set bits [OK]

Hint: Clear least set bit repeatedly to count set bits [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Using string conversion instead of bit operations
• Shifting without checking bits

2. Which of the following Rust code snippets correctly implements the Brian Kernighan algorithm to count set bits?

easy

A. fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n > 0 { count += n & 1; n >>= 1; } count }

B. fn count_set_bits(n: u32) -> u32 { n.count_ones() }

C. fn count_set_bits(n: u32) -> u32 { n.trailing_zeros() }

D. fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count }

Solution

1. Step 1: Identify the Brian Kernighan algorithm code

   fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count } uses the loop with n &= n - 1 to clear the least significant set bit repeatedly.

2. Step 2: Verify other options

   fn count_set_bits(n: u32) -> u32 { n.count_ones() } uses built-in count_ones(), fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n > 0 { count += n & 1; n >>= 1; } count } uses bit shifting, and fn count_set_bits(n: u32) -> u32 { n.trailing_zeros() } uses trailing_zeros(), which are not Brian Kernighan's method.

3. Final Answer:

   fn count_set_bits(mut n: u32) -> u32 { let mut count = 0; while n != 0 { n &= n - 1; count += 1; } count } -> Option D

4. Quick Check:

   n &= n - 1 loop is Brian Kernighan's method [OK]

Hint: Use n &= n - 1 in loop to clear bits [OK]

Common Mistakes:

• Using built-in functions instead of algorithm
• Shifting bits instead of clearing
• Using trailing_zeros which counts zeros, not ones

3. What is the output of the following Rust code?

fn main() {
    let mut n = 29u32; // binary 11101
    let mut count = 0;
    while n != 0 {
        n &= n - 1;
        count += 1;
    }
    println!("{}", count);
}

medium

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Convert 29 to binary and count set bits

   29 in binary is 11101, which has four set bits (1s).

2. Step 2: Apply Brian Kernighan algorithm steps

   Each iteration clears one set bit until n becomes zero, so count increments 4 times.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Number of 1s in 11101 = 4 [OK]

Hint: Count 1s in binary representation [OK]

Common Mistakes:

• Counting total bits instead of set bits
• Miscounting binary digits
• Confusing zero bits with set bits

4. Identify the error in this Rust implementation of Brian Kernighan's algorithm:

fn count_set_bits(mut n: u32) -> u32 {
    let mut count = 0;
    while n > 0 {
        n = n - 1;
        count += 1;
    }
    count
}

medium

A. It uses n > 0 instead of n != 0 in the loop condition

B. It uses a while loop instead of a for loop

C. It decrements n by 1 instead of clearing the least significant set bit

D. It does not initialize count to zero

Solution

1. Step 1: Analyze the loop operation on n

   The code subtracts 1 from n each time, which does not clear the least significant set bit correctly.

2. Step 2: Understand Brian Kernighan's method

   The correct operation is n &= n - 1 to clear the least significant set bit, not n = n - 1.

3. Final Answer:

   It decrements n by 1 instead of clearing the least significant set bit -> Option C

4. Quick Check:

   Use n &= n - 1 to clear bits, not n = n - 1 [OK]

Hint: Use bitwise AND to clear bits, not subtraction alone [OK]

Common Mistakes:

• Using subtraction instead of bitwise AND
• Wrong loop condition causing infinite loop
• Not incrementing count properly

5. Given a list of unsigned integers [7, 8, 15, 23], which Rust code using Brian Kernighan's algorithm correctly returns a vector of their set bit counts?

hard

A. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() }

B. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.count_ones()).collect() }

C. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.trailing_zeros()).collect() }

D. fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| n.leading_zeros()).collect() }

Solution

1. Step 1: Identify Brian Kernighan's algorithm in code

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } uses the loop with x &= x - 1 to count set bits for each number.

2. Step 2: Check other options

   Options B, C, and D use built-in functions not related to Brian Kernighan's algorithm.

3. Step 3: Confirm output correctness

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } returns a vector with counts of set bits for each number in the list.

4. Final Answer:

   fn count_bits_list(nums: &[u32]) -> Vec<u32> { nums.iter().map(|&n| { let mut x = n; let mut c = 0; while x != 0 { x &= x - 1; c += 1; } c }).collect() } -> Option A

5. Quick Check:

   Loop with x &= x - 1 counts bits per number [OK]

Hint: Map each number with Brian Kernighan loop to count bits [OK]

Common Mistakes:

• Using built-in count_ones instead of algorithm
• Using trailing_zeros or leading_zeros which count zeros
• Not collecting results into a vector